JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Three Dimensional Geometry

Let $Q$ be the foot of the perpendicular from $P(1,6,a)$ to $L$; since $P^{\prime }\left(\frac{a}{3},0,a+c\right)$ is the image of $P$, $Q$ is the midpoint of $P{P}^{\prime }$. A point of $L$ is $(k,2k+1,bk+a-1)$ with direction $(1,2,b)$. Requiring this point to be the foot, with $\overrightarrow{PQ}\perp (1,2,b)$, together with the midpoint coordinates gives $a=3$, and with $b>0$, $b=3$ and $c=4$; then $Q=(1,3,5)$. A general point of $L$ is $(k,2k+1,3k+2)$, and $SQ=2\sqrt{14}$ gives $k=3$ or $k=-1$. Since $\alpha >0$, $k=3$, so $S=(3,7,11)$ and $\alpha +\beta +\gamma =3+7+11=21$.